泰勒系列(麦克劳伦)，后9或11，编译器写；

Question

泰勒级数中sin(x)的公式(代码下的图片)。通常，如果在步骤2中输入Start 1和end 20，则控制台在x=9之后输出'-nan‘；sin(X)和Taylor应该是相同的；例如：

X= 9；sin(x) = 0.412118；taylor = 0.412118

X= 11；sin(x) = -0.99999；taylor -0.999976；

X= 13；sin(x) = 0.420167；taylor = -nan；

一直这样，我需要一些帮助，这是为了我的实验室

#include <stdio.h>
#include <math.h>

int main(void) {
    float a, b, left, right, eps = 0.00001, step, x, add = 1, chis, znam, fact, sum = 0, delta;
    printf("Plese enter your start: ");
    scanf("%f", &a);
    printf("Your end: ");
    scanf("%f", &b);
    printf("and step: ");
    scanf("%f", &step);

    if (b < a || a < eps) {
        printf("Your inputs aren't correct");
        return -1;
    }

    printf("\tX\t           sin(x)\tTaylor\t   Delta\n");
    for (x = a; x < b; x += step) {

        printf("    x = %9f\t", x);
        left = sin(x);
        printf("%9f", left);

        chis = -x;
        znam = 1;

        sum = 0;
        add = 1;
        fact = 1;
        while (fabs(add) > eps) {
            add = -1 * chis / znam;
            sum += add;
            chis *= -1 * (x * x);
            fact++;
            znam *= fact * (fact + 1);
            fact++;

        }
        printf("    %9f    ", sum);
        printf("%e\n", fabs(left - sum));
    }
}

​

​

Answer 1

您不需要计算幂或阶乘：

double tailor(double x, unsigned n)
{
    double result = x;
    double part = x;
    int sign = -1;

    for(unsigned i = 1; i <= n; i++)
    {
        for(unsigned dv = 2 * (i - 1) + 1 + 1; dv <= 2 * i + 1; dv++)
        {
            part *= x / (double)dv;
        }        
        result += sign * part;
        sign *= -1;
    }
    return result;
}


int main(void)
{
    for(unsigned n = 10; n < 60; n++)
        printf("n = %u : %.60f\n", n, tailor(13, n));
}

https://godbolt.org/z/b1z3rszh8

您可以很容易地添加epsilon检查，而不是从调用方填充n。